Question

The sum of first 6th term is 42 and the ratio of its 10th term to its 30th term is1:3 caculate the 1st and 13 term of an ap

Answer 1

s6 = 42
n/2[2a+(n-1)d] = 42
6/2[2a+5d]=42
3(2a+5d) = 42
2a+5d = 14
a=14-5d/2 ------1
s10:s30
=10/2(2a+9d):30/2(2a+29d)
=5(2a+9d):15(2a+29d)
=2a+9d:6a+87d
=0:4a+78d
=0:2a+39d
from eq 1 put the value of a
=0:2[14-5d/2]+39d
=0:14-5d+39d
=0:14+34d
=d=-1/2
put in eq 1
a=14-5d/2
a=(14+5/2)/2
a=33/2
a13=a+(n-1)d
a13=33/2+12(-1/2)
a13=(33-12)/2
a13=21/2